Chemical, Biological and Environmental Engineering
Cost of Electrical Energy
AC Power Concepts
Advanced Materials and Sustainable Energy LabCBEE
Housekeeping IssuesChanged way #2 is stated (replaced n2 with T2 for clarity)
Timeline: Midterm planned for Tuesday 2/8 (date OK?)
Class on 2/3: Intro to Nuclear Energy
Plan to issue HW4 on 2/1(due on 2/8 before midterm)
GS: Need to discuss class presentation subject
Reminder: visit to Energy center on 2/1 at class time.
Advanced Materials and Sustainable Energy LabCBEE
Advanced Materials and Sustainable Energy LabCBEE
Baseload, Intermediate and Peaking Supply
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Baseload, Intermediate and Peaking Supply• Load demand on utilities fluctuates constantly
– During peak demand most plants are operating– During light demand many plants are idling
• Power plants are categorized as– Baseload
• Large coal, nuclear, and hydroelectric plants• Expensive to build, cheap to operate
– Intermediate• Combined-cycle plants• Cycled up during the day, cycled down during the evening
– Peaking• Simple-cycle gas turbines• Inexpensive to build, expensive to operate
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Costing Power• Other fixed costs
– Regular maintenance (e.g., groundskeeping)– Administration– Insurance
• Variable Costs (primarily fuel) – Cost of fuel
• Coal ~ $2.21/MMBTU ($43.74/ton)• Gas ~ $4.74/MMBTU
– Operations and Maintenance (Repair & Spare parts, etc)
Variable Costs ($/yr) =
[ Fuel($/BTU)xHeat rate(BTU/kWh) + O&M($/kWh)] xkWh/yr
Advanced Materials and Sustainable Energy LabCBEE
Costing Power - II
Total cost of operating power plant then is the sum:
Then, depending on how many kWh are generated in a typical year,
This levelized cost per unit of energy is useful to compare various projects
Total Annual Cost [$ / ]Levelized Cost [$ / ] $ /
Annual Output [ / ]
yrkWh kWh
kWh yr= =
Total Annual Cost [$ / ]
Total Fixed Costs [$ / ] Total Variable Costs [$ / ]
yr
yr yr
=
= +
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Graphical Version of Costing Power
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New Generation Costs SummaryCapital Costs ($/kW)
Fixed O&M ($/kW)
Heat Rate (BTU/kWhr)
Variable O&M(¢/kWh)
Conventional Coal 2,200 28 8,700 2.4
IGCC (Integrated Coal Gasification Combined Cycle)
2,600 40 7,500 2.0
IGCC with CCS 3,800 47 8,300 2.3
Gas Combined Cycle 1,000 12 6,300 3.2
Gas CC w/CCS 2,000 20 7,500 3.9
Gas Turbine 650 11 10,500 5.3
Nuclear 3,800 92 10,500 1.2
Wind 2,000 31 - -
Wind-Offshore 4,000 87 - -
Hydro 2,300 14 - 0.2
Geothermal (US average) 1,750 168 30,000 0.5
Solar PV 6,200 12 - -
Solar Thermal 5,100 58 - -
Adapted from EIA publicationElectricity Market Module of the National Energy Modeling System 2010, DOE/EIA-M068(2010)
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Costing Power
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Screening Curve• Plot costs for different plants on the same graph
Plot as Cost = Fixed Cost ($) + Variable Cost ($/kWh) *kWh
Cost
($)
Energy Produced (kWh) [or rated power (kW) x hours of operation (h)]
Fixed Costs Plant 1 ($)
Variable Costs Plant 1 ($/kWh * kWh)
Fixed Costs Plant 2Variable Costs Plant 2
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Using Screening CurvesCombustion turbine is lowest-cost option for up to 1675 h/yr of operationCoal plant is the lowest-cost option for operation beyond 6565 h/yrThe combined cycle plant is the cheapest option if it runs between 1675 and
6565 h/yr
Advanced Materials and Sustainable Energy LabCBEE
Capacity FactorThe capacity factor is defined as
CF = [produced energy per year (kWh/yr)] / [ Rated power (kW) x 8760 h/yr]
Essentially “fraction of plant on-line time at full power averaged over year”
Why would the plant not operate at full rated power for full year?1-Time down for maintenance (try to minimize this…)2-Power it produces is not cost effective (use screening curve to find out how many hours on-line)
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Load-Duration Curves
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Load-Duration Curve
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Determining Optimum MixTransfer crossover points onto load duration curve to identify
optimum mix of power plants
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LOAD-DURATION CURVES
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Determining Optimum Mix
Baseload
IntermediatePeaking
+ Reserve
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Baseload, Intermediate and Peaking Supply
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Cost of PowerThe CF with cost parameters allow us to determine the cost of
electricity from each type of plant
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Cost of Energy to UserThe final price is the weighted average of the costs of
all generation:
From example on previous slide:
Final Cost [ / ] /Output UnitCost
c kWh c kWhOutput
´/ /= =å
å
9
9
Final Cost [ / ]
27.99[10 ] 4.69[ / ] 4.94 6.23 1.14 12.87
27.99[10 ] 4.94 1.14
5.19 /
c kWh
kWh c kWh x x
kWh
c kWh
/ =
/´ + += =
+ +
/=
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Costing Power• Load duration curve needs to be padded with
reserve excess capacity (reserve margin) – To deal with plant outages, sudden peaks in demand, and
other unforeseen events
• Process of selecting which plant to operate first at any given time is called dispatching
• If you have renewables, they will be dispatched first, although they are intermittent (and require extra spinning reserve)– Energy Policy Act 1992/2005
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New Generation by Fuel Type(USA 1990 to 2030, GW)
Source: EIA Annual Energy Outlook 2007
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Upcoming Challenges to GridPlug in Electric Vehicles: What do you think that will do to the daily load profile?
People come home at about this time
In absence of incentiveto charge at other timepeak will get higher (worse)
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Smart GridQ: Why should someone using power off peak pay rates using
peak power? (“subsidizing peak power”?)
Adding information layer to power distribution grid• Allows utilities to charge differential rates depending on mix• Allows utilities to inform customers about changing rates• Allows customers to decide when to consume power• Can also be used (voluntarily) to turn off non-essential loads at
customers– Pool Pumps, Washer/Drier, A/C– “Demand Side Management”
Also• Increases ability of users to add distributed generation• Increases resilience of grid
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Example of Distributed Generation: Microturbines
Very small gas turbines (“aeroderived”)
• 1kW to several 100 kW• 30 to 60 kW unit about the size of a refrigerator• Have only one moving part
– Spins at ~96000 rpm on air bearings…
• Particularly good for CHP
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MICROTURBINESGenerator makes variable frequency AC that is rectified and
inverted to grid frequency ac (50 or 60 Hz)
Some microturbines are designed for power and heat
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MICROTURBINES
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Microturbines
Source: http://www.capstoneturbine.com
Capstone 65 kW Microturbine
230 kW fuel
80% CHP Heat Utilization
120 kW hot water
65 kW electrical
45 kW waste heat
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DG/CHP: Fuel cellsExample: ClearEdge Power (Hillsboro – OR!)• PEM fuel cell operating on natural gas (reformate)• No moving parts (but membrane degrades)• ~40% Efficiency, ~85% Energy Utilization
Chemical, Biological and Environmental Engineering
Evaluating power consumption in AC
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AC Power
Instantaneous Power = VI
What is the power transferred if V=120V, I=10A,
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AC Power
cos sin and 2j te t j t fw w w w p= + =
Note: because i is also used for current, the imaginary number basis “i” is usually replaced by “j”
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Phasor Representation
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AC Power
Previous problem reduces to: “What is the instantaneous power transferred if V=120V, I=10A and f=0”
What if f=90o instead?
P=VIcosf
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Real vs. Apparent PowerReal power is the capacity of the circuit for performing
work in a particular time
Apparent power is the product of the current and voltage of the circuit
If Voltage and Current out of phase, apparent power can be greater than real power
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Real, Apparent, Reactive Power; Power Factor
Real Power usually written as P, given in Watts
Apparent Power is S, Given in Volt-Amperes(i,.e., based on system voltage and current)
The difference is “Reactive Power”, given as Q Basically, current flowing 90o (π/2) out of phase with voltage
Given in Volt-Amperes Reactive (or VARs)
S can be calculated as S2=P2+Q2
Power factor l=P/S (fraction of power that is useful)Can be shown that P=S cosf, therefore l=cosf
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About the Power FactorA load with low power factor draws more current than
a load with a high power factor for the same amount of useful power transferred
Higher currents increase losses in the distribution system, and require larger wires and other equipment
Causes utilities to charge higher cost for low power factor – it is in your interest to correct power factor
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AC Power
Further, power varies continuously
what is the power at the point V (or I) crosses 0?
Use “Root Mean Squared” (RMS) power instead
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Root Mean Squared Values
2
By definition
For a sinusoid
sin(2 ) you can show that 2
, 2 2
and cos cos2
rms avg
rms
m mrms rms
m mrms rms
A A
aA a ft A
Therefore
V IV I
V IP V I
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ImpedanceIn general you know that power is dissipated by
a current flowing against a voltage (V=IR)
In AC you have to consider the effect of sinusoidal voltage waveform
The generic “resistance is called “impedance”– Purely resistive (R)– Capacitive (XC)
– Inductive (XL)
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Complex Impedance• In general impedance is called “Z”
– “R” is still the resistance– “X” is the “reactance”
• The relationship is
jVZ e R jX
If-= = +
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Inductive Reactance• We define a quantity called the Inductance, L
• From Faraday’s law
NL
i
F=
d diN iL N L
dt dt
FÛ F = Þ =
dN V
dte
F= =
di diL V LV
dt dtÛ = Û =
0
1( ) ( ) (0)
t
i t V t dt iL
= +ò
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Inductive ReactanceIn other words, current lags voltage change
• Units of inductance is the “Henry” written as H
and LL
VI X L
Xw= =
(and 2 )fw p=
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Capacitive reactanceHow does capacitor work?
Current flows onto capacitor electrodes
Charge accumulation builds voltage
This causes current to lead voltage
Impedance of a capacitor is called capacitance measured in farads
dqC
dV=
( ) ( )( )
dq t dV ti t C
dt dt= =
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Capacitive reactance• Therefore current leads voltage in a capacitor
1 and C
C
VI X
X Cw= =
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In Phasor Diagrams…
( )22L CZ R X X= + -
c
tan L CX X
Rf
-=1tan L CX X
Rf - æ ö- ÷ç= ÷ç ÷÷çè ø
L CX X X= -
( )22 2L CZ R X X= + -
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You can combine impedances• In Series…
• In Parallel…
• Etc…
1 2eqZ Z Z= +
1 2
1 2 1 2
1 1 1eq
eq
Z ZZ
Z Z Z Z Z= + Û =
+
2 31 2 3 1
3 3eq
Z ZZ Z Z Z Z
Z Z= + = +
+